No Solution for 3x+2y=7 in Real Numbers

[relyt]

Statement:

(\exists_{x} \in R) (\forall_{y} \in R) (3x + 2y = 7)

Trying to find negation statement. This is what I think it is:

(\forall_{x} \in R)(\exists_{y} \in R) (3x + 2y ≠ 7)


Is this close?

 

[phreak]

I'm having trouble reading this notation, but I'm assuming the first statement says "There exists an x in R such that for all y in R, 3x + 2y = 7."

To make this backwards, we would need to say, "For each x in R, there exists a y in R such that 3x + 2y does not equal 7," so if I have interpreted your notation correctly, I think your answer is correct.

 

[relyt]

Thanks, phreak. Yes, that is how it should read.